Transcript Right Triangle Problems - Northland Preparatory Academy

Sec. 4.8
New Definitions
Angle of Elevation – angle through which the eye moves up
from horizontal to look at something above.
Angle of Depression – angle through which the eye moves down
from horizontal to look at something below.
Consider two people looking at each other from the top
and bottom of a flight of stairs:
Angle of
Depression
Angle of
Elevation
Guided Practice
The angle of depression of a buoy from the top of a lighthouse
130 feet above the surface of the water is 6 . Find the distance
x from the base of the lighthouse to the buoy.
First, draw a diagram:
Solve for x algebraically:
6
130 ft
x
6
130
tan 6 
x
130
x
tan 6
x  1236.867 feet
Guided Practice
100 ft
From the top of a 100-ft tall building a man observes a car moving
toward the building. If the angle of depression of the car changes
from 22 to 46 during the period of observation, how far does the
car travel?
The diagram:
22
46
22
x
46
d
From the smaller right triangle:
100
100
tan 46 
d 
d
tan 46
Guided Practice
100 ft
From the top of a 100-ft tall building a man observes a car moving
toward the building. If the angle of depression of the car changes
from 22 to 46 during the period of observation, how far does the
car travel?
The diagram:
22
46
22
46
x
d
From the larger right triangle:
100
tan 22 
xd
100
d
tan 46
Substitute!!!
100
100
xd 
x
d
tan 22
tan 22
Guided Practice
100 ft
From the top of a 100-ft tall building a man observes a car moving
toward the building. If the angle of depression of the car changes
from 22 to 46 during the period of observation, how far does the
car travel?
The diagram:
22
46
22
x
46
d
100
100
 150.940 feet
x

tan 22 tan 46
Guided Practice
A large, helium-filled penguin is moored at the beginning of a
parade route. Two cables attached to the underside of the
penguin make angles of 48 and 40 with the ground and are in
the same plane as a perpendicular line from the penguin to the
ground. If the cables are attached to the ground 10 feet from
each other, how high above the ground is the penguin?
Definition of the tangent function:
The diagram:
h
h
tan 48 
tan 40 
x
x  10
h
40
10
48
x
Solve both for h:
h  x tan 48 h   x  10 tan 40
Set equal, solve for x:
x tan 48   x 10 tan 40
Guided Practice
A large, helium-filled penguin is moored at the beginning of a
parade route. Two cables attached to the underside of the
penguin make angles of 48 and 40 with the ground and are in
the same plane as a perpendicular line from the penguin to the
ground. If the cables are attached to the ground 10 feet from
each other, how high above the ground is the penguin?
x tan 48   x 10 tan 40
x tan 48  x tan 40  10 tan 40
The diagram:
x tan 48  x tan 40  10 tan 40
h
40
10
48
x
x  tan 48  tan 40  10tan 40
10 tan 40
x
tan 48  tan 40
Plug back in to solve for h!!!
Guided Practice
A large, helium-filled penguin is moored at the beginning of a
parade route. Two cables attached to the underside of the
penguin make angles of 48 and 40 with the ground and are in
the same plane as a perpendicular line from the penguin to the
ground. If the cables are attached to the ground 10 feet from
each other, how high above the ground is the penguin?
h  x tan 48
The diagram:
h
40
10
10 tan 40 tan 48
h
tan 48  tan 40
 34.323 feet
48
x
Guided Practice
The top row of the red seats behind home plate at Cincinnati’s
Riverfront Stadium is 90ft above the level of the playing field.
The angle of depression to the base of the left field wall is 14 .
How far is the base of the left field wall from a point on level
ground directly below the top row?
d = distance in question
90
d
 90 cot14  360.970 feet
tan14
Guided Practice
The angle of elevation of the top of Proctor’s ego is measured to
be 441357 at a point 12.3 feet from the base of the ego. How
tall is Proctor’s ego?
h = height of Proctor’s ego
13   57 

441357  44      
   44.2325
 60   3600 
h  12.3 tan 44.2325  11.975 feet
Guided Practice
While hiking on a level path toward Colorado’s front range, Otis
Evans determines that the angle of elevation to the top of Long’s
Peak is 30 . Moving 1000 ft closer to the mountain, Otis
determines the angle of elevation to be 35 . How much higher is
the top of Long’s Peak than Otis’s elevation?
h = height in question
1000
h
 3290.526
cot 30  cot 35
feet
A U.S. Coast Guard patrol boat leaves Port Cleveland and
averages 35 knots (nautical mph) traveling for 2 hours on a course
of 53 and then 3 hours on a course of 143 . What is the boat’s
bearing and distance from Port Cleveland?
First, let’s diagram the path of the boat…
Now, we need the measure of angle ABC…
B 143
70
53
53 37
How about the distances AB and BC?
Distance = Speed x Time
105
AB = (35 knots)(2 hours)
= 70 naut. mi.
A
BC = (35 knots)(3 hours)
= 105 naut. mi.
C
A U.S. Coast Guard patrol boat leaves Port Cleveland and
averages 35 knots (nautical mph) traveling for 2 hours on a course
of 53 and then 3 hours on a course of 143 . What is the boat’s
bearing and distance from Port Cleveland?
Solve the right triangle for AC and  :
B 143
70
53
A

53 37
AC  70 105
 126.194 naut. mi.
105
tan  
70
1  105 


tan


C
 70 
 56.310
2
105
2
A U.S. Coast Guard patrol boat leaves Port Cleveland and
averages 35 knots (nautical mph) traveling for 2 hours on a course
of 53 and then 3 hours on a course of 143 . What is the boat’s
bearing and distance from Port Cleveland?
Interpret our answer:
B 143
70
53
A

53 37
The patrol boat is about 126
nautical miles from port, at a
bearing of approximately
105 109.3 .
C
More Practice Problems
A boat travels at 40 knots from its home port on a course of 65
for 2 hours and then changes to a course of 155 for 4 hours.
Find the distance and bearing from the port to the boat.
155
65

80
160
d  80 160
 178.885 nautical miles
Port
2
tan   2   tan
d
Bearing
2
 tan
1
1
 2
 2  65  128.435
More Practice Problems
A ranger spots a fire from a 73-ft tower in Yellowstone National
Park. She measures the angle of depression to be 120. How
far is the fire from the tower?
 4 3 
120   4 3 
73
 4 3 
d
73
tan  4 3  
d
d  73cot  4 3   3136.378 ft
More Practice Problems
A six meter ladder makes an angle of 70 with the ground as it
leans against a wall. The ladder then slips slightly so that the
angle it makes with the ground changes to 60 . How much
higher on the wall was the ladder before it slipped?
Before slipping:
After slipping:
x
sin 70 
6
6m
70
x
y
sin 60 
6
6m
60
y
More Practice Problems
A six meter ladder makes an angle of 70 with the ground as it
leans against a wall. The ladder then slips slightly so that the
angle it makes with the ground changes to 60 . How much
higher on the wall was the ladder before it slipped?
x
sin 70 
6
y
sin 60 
6
x  6 sin 70 y  6sin 60
Total change in vertical position on wall:
x  y  6sin 70  6sin 60
 0.442 meters